By Soo Jin Lee and Jaehong Shin
1. x(n+1) = 3.2[x(n)][1-x(n)]
Consider the behavior of the sequence for various x(0). Use a spreadsheet to display a table and graph for a sufficient range of n.
In particular, investigate the behavior of the sequence for x(0) in the range of x(0) = 0.6875
Through a few trial and error, I came to know that I could narrow down the range of x(0) for investigation to 0<x(0)<1 since the sequences has the value 0 when x(0) = 1 or 0, otherwise sequences goes to the negative infinity as follows.
Therefore, I observed the behavior of the sequence
for various x(0): from 0.1 to 0.5 at regular 0.1 interval. Also
I came to realized that I did not have to explore values between
0.5 and 1 because x(0) and 1-x(0) has the same sequences by the
suggested relation of sequences except the value of x(0).
As we can see above, although the starting point is different each other, it seems to be focused on two same values while oscillating between two values on its own: 0.79945549 and 0.51304451.
By the way, when I input 0.6875 for the value of x(0), I observed very peculiar features.
All values of the sequences are same as the first value, that is, 0.6875. To find another possibility to show such a phenomenon, I tested for 0.6874 and 0.6876.
In those cases, the sequence value is focused on the same value: 0.79945549 and 0.51304451. Through previous explorations, I could guess 0.6875 is the only value that makes all sequences the same value as itself. Then how can I explain that conjecture mathematically?
Explanation) I made a quadratic equation under
the assumption that x(n+1) = x(n).
x(n) = 3.2[x(n)][1-x(n)]
_ 3.2[x(n)2] 2.2x(n) = 0
_ x = 0 or 2.2/3.2
Therefore, 2.2/3.2 = 0.6875 is the only solution between 0 and 1.
For exploration using excel: click!
2. Investigate this sequence for a range
x(n+1) = 3.83[x(n)][1-x(n)]
In this case, also I narrowed down the range of x(0) for investigation to 0<x(0)<1 for the same reason as problem 1. The result was also similar. Regardless of the value of x(0), all sequences are focused on three values: 0.156149316, 0.504666487 and 0.957416598.
Now, can I get the value that makes all sequence
values equal in this time? By the same manner in exploration 1,
I got the value 2.83/3.83. However, the result is somewhat different.
At first, it seemed to have same values but
as n increased, sequence values went to the previous three values:
0.156149316, 0.504666487 and 0.957416598. I am sure the solution
2.83/3.83 is mathematically right. Then what is the problem? I
concluded that the cause is the error of spreadsheet program.
In fact, when we calculate 2.83/3.83 we can not get a finite decimal
contrary to the finite decimal 2.2/3.2 = 0.6875. However, Microsoft
Excel program that I used to explore supports only 15digit calculation
and provides an approximate value. The error resulting from the
approximation made the sequence to diverge. Therefore, we can
guess that it is impossible to make a correct constant sequence
using computer program because computer is certain to approximate
at the end.
For exploration using excel: click!!
3. For 0<A<2 examine x(n+1) = 1
To begin with, I fixed the value of x(0) as 0.1
At least up to 0.7 all sequences converges to one value although the converging values are different from each other. However, as A is greater 0.7, the shape of a graph started to differ.
As we can see above, gradually the focused point is separated by two although we can not say that it is focused on two points rigorously because the upper value is decreasing little by little by a very small quantity and the lower value sequences are increasing also.
On the other hand, for various x(n), it showed
a similar phases, that is, for each A value the sequence intend
to go to the same value regardless of the value of x(n).